「CC PRIMEDST」Prime Distance On Tree-点分治 + FFT

给一棵树,随机选取两个点,求两点间路径距离为质数的概率。

链接

CC PRIMEDST

题解

用点分治来统计个数,发现求的是
$$\sum_{i + j = P}cnt_i \cdot cnt_j$$
这是一个卷积的形式,于是 FFT 就好了(要爆 int,还是别写 NTT 了…)

时间复杂度 $O(n \log ^ 2 n)$。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
/**
* Copyright (c) 2016-2018, xehoth
* All rights reserved.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
* http://www.apache.org/licenses/LICENSE-2.0
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
* 「CC PRIMEDST」Prime Distance On Tree 02-05-2018
* 点分治 + FFT
* @author xehoth
*/
#include <bits/stdc++.h>
struct InputStream {
enum { SIZE = 1024 * 1024 };
char ibuf[SIZE], *s, *t, obuf[SIZE], *oh;
inline char read() {
if (s == t) t = (s = ibuf) + fread(ibuf, 1, SIZE, stdin);
return s == t ? -1 : *s++;
}
template <typename T>
inline InputStream &operator>>(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return *this;
iosig |= c == '-';
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
if (iosig) x = -x;
return *this;
}
} io;
constexpr double PI = acos(-1);
const int MAXN = 50000;
const int MAXM = 1 << 17 | 1;
struct Complex {
double r, i;
inline Complex operator+(const Complex &p) const {
return {r + p.r, i + p.i};
}
inline Complex operator-(const Complex &p) const {
return {r - p.r, i - p.i};
}
inline Complex operator*(const Complex &p) const {
return {r * p.r - i * p.i, r * p.i + i * p.r};
}
inline Complex conj() const { return {r, -i}; }
inline Complex inv() const {
double x = r * r + i * i;
return {r / x, -i / x};
}
inline void operator+=(const Complex &p) {
r += p.r;
i += p.i;
}
} rt[MAXM], irt[MAXM], a[MAXM];
inline void init(int n) {
rt[0] = {1, 0};
rt[1] = {cos(2 * PI / n / 2), sin(2 * PI / n / 2)};
for (int i = 2; i < n; i++) rt[i] = rt[i - 1] * rt[1];
irt[0] = {1, 0};
irt[1] = rt[1].inv();
for (int i = 2; i < n; i++) irt[i] = irt[i - 1] * irt[1];
for (int i = 0, j = 0; i < n; i++) {
if (i > j) {
std::swap(rt[i], rt[j]);
std::swap(irt[i], irt[j]);
}
for (int k = n >> 1; (j ^= k) < k; k >>= 1)
;
}
}
inline void dit(Complex *a, int n) {
for (int i = 1, l = n >> 1; i < n; i <<= 1, l >>= 1) {
for (int j = 0, o = 0; j < i; j++, o += l << 1) {
const Complex &w = rt[i + j];
for (int k = o; k < o + l; k++) {
Complex t = a[k + l] * w;
a[k + l] = a[k] - t;
a[k] += t;
}
}
}
}
inline void dif(Complex *a, int n) {
for (int i = n >> 1, l = 1; i; i >>= 1, l <<= 1) {
for (int j = 0, o = 0; j < i; j++, o += l << 1) {
const Complex &w = irt[i + j];
for (int k = o; k < o + l; k++) {
Complex t = a[k + l];
a[k + l] = (a[k] - t) * w;
a[k] += t;
}
}
}
for (int i = 0; i < n; i++) a[i].r /= n;
}
int pr[MAXN + 1], cnt, n;
bool isV[MAXM * 2 + 9];
inline void sieve(int n) {
isV[1] = true;
for (int i = 2; i <= n; i++) {
if (!isV[i]) pr[cnt++] = i;
for (int j = 0, k; j < cnt && (k = i * pr[j]) <= n; j++) {
isV[k] = true;
if (i % pr[j] == 0) break;
}
}
}
std::vector<int> g[MAXN + 9];
long long ans;
int sz[MAXN + 9];
bool vis[MAXN + 9];
void dfsSize(int u, int pre) {
sz[u] = 1;
for (int v : g[u]) {
if (!vis[v] && v != pre) {
dfsSize(v, u);
sz[u] += sz[v];
}
}
}
int get(int u, int pre, int n) {
for (int v : g[u])
if (!vis[v] && v != pre && sz[v] > n) return get(v, u, n);
return u;
}
std::vector<long long> sub, all;
void dfsDep(int u, int pre, int dep) {
if (!isV[dep]) ans += 2;
for (; (int)sub.size() <= dep;) sub.push_back(0);
sub[dep]++;
for (; (int)all.size() <= dep;) all.push_back(0);
all[dep]++;
for (int v : g[u])
if (!vis[v] && v != pre) dfsDep(v, u, dep + 1);
}
inline void convolve(std::vector<long long> &v) {
int m = v.size() + v.size() + 1, k = 1;
for (; k <= m;) k <<= 1;
for (int i = 0; i < (int)v.size(); i++) {
a[i].r = v[i];
a[i].i = 0;
}
for (int i = (int)v.size(); i < k; i++) a[i].r = a[i].i = 0;
dit(a, k);
for (int i = 0; i < k; i++) a[i] = a[i] * a[i];
dif(a, k);
for (; (int)(a[m].r + 0.3) == 0 && m;) m--;
v.resize(m + 1);
for (int i = 0; i <= m; i++) v[i] = (long long)(a[i].r + 0.3);
}
void solve(int u) {
dfsSize(u, 0);
vis[u = get(u, 0, sz[u] / 2)] = true;
all.clear();
for (int v : g[u]) {
if (!vis[v]) {
sub.clear();
dfsDep(v, u, 1);
convolve(sub);
for (int j = 0; j < cnt && pr[j] < (int)sub.size(); j++)
ans -= sub[pr[j]];
}
}
convolve(all);
for (int j = 0; j < cnt && pr[j] < (int)all.size(); j++) {
ans += all[pr[j]];
}
for (int v : g[u])
if (!vis[v]) solve(v);
}
int main() {
// freopen("sample/1.in", "r", stdin);
sieve(MAXN * 2 + 1);
io >> n;
for (int i = 1, u, v; i < n; i++) {
io >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
for (int i = 1; i <= n; i++) std::reverse(g[i].begin(), g[i].end());
int k = 1;
for (; k <= n + n + 1;) k <<= 1;
init(k);
solve(1);
printf("%.8f", (double)ans / ((double)n * (n - 1) / 2.0) / 2.0);
return 0;
}
分享到